Problem: What is the average value of $\sqrt[3]{x}$ on the interval $-5\leq x \leq 9$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{3}{56}\cdot(\sqrt[4]{9^3}-\sqrt[4]{5^3})$ (Choice B) B $\dfrac{3}{56}\cdot(\sqrt[3]{9^4}-\sqrt[3]{5^4})$ (Choice C) C $\dfrac{1}{2}\cdot (\sqrt[3]{9}-\sqrt[3]{-5})$ (Choice D) D $\dfrac{1}{2}\cdot (\sqrt[3]{9}+\sqrt[3]{-5})$
Explanation: In general, this is the average value of function $f$ over the interval $[a,b]$ : $\dfrac{\int_a^b f(x)\,dx}{b-a}$ In our case, ${f(x)=\sqrt[3]{x}}$, ${a=-5}$ and ${b=9}$ : $\begin{aligned} \dfrac{\int_{ a}^{ b} {f(x)}\,dx}{ b- a}&=\dfrac{\int_{{-5}}^{ {9}} ({\sqrt[3]{x}})\,dx}{{9}-{(-5)}} \\\\ &=\dfrac{\left[\dfrac{3\cdot x^{^{\frac{4}{3}}}}{4}\right]_{-5}^{9}}{14} \\\\ &=\dfrac{\dfrac{3\cdot 9^{^{\frac{4}{3}}}}{4}-\dfrac{3\cdot (-5)^{^{\frac{4}{3}}}}{4}}{14} \\\\ &=\dfrac{3}{56}\cdot(\sqrt[3]{9^4}-\sqrt[3]{5^4}) \end{aligned}$ In conclusion, the average value of $\sqrt[3]{x}$ on the interval $-5\leq x \leq 9$ is $\dfrac{3}{56}\cdot(\sqrt[3]{9^4}-\sqrt[3]{5^4})$.